Also available at http://amc.imfm.siISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)

ARS MATHEMATICA CONTEMPORANEA 3 (2010) 109–120

Fullerene patches I

Jack E. GraverSyracuse University, Department of Mathematics, Syracuse, NY 13244-1150, USA

Christina M. GravesThe University of Texas at Tyler, Department of Mathematics, Tyler, TX 75799, USA

Received 10 December 2009, accepted 17 April 2010, published online 10 May 2010

Abstract

Large carbon molecules, discovered at the end of the last century, are called fullerenes.The most famous of these, C60 has the structure of the soccer ball: the seams representchemical bonds and the points where three seams come together represent the carbonatoms. We use the term fullerene to represent all mathematically possible structures forthe chemical fullerenes: trivalent plane graphs with only hexagonal and pentagonal faces.A simple consequence of Euler’s formula is that each fullerene has exactly 12 pentago-nal faces; the only restriction on the number of hexagonal faces is that it not be 1. Thechemical motivation for this paper is to answer the question: When is it possible to altera fullerene by changing the structure inside a region of the fullerene bounded by a simpleclosed circuit? Such a region or patch is said to be ambiguous if alterations may be madeto its interior without disturbing the structure of the fullerene outside of the region. In thispaper, we show that, relative to the minimum distance between pentagonal faces, there areno small ambiguous patches.

Keywords: Fullerenes, fullerene patches, ambiguous patches, graphite patches.

Math. Subj. Class.: 05C10, 05C75, 92E10

1 IntroductionLet Γ = (V,E, F ) be a fullerene, a trivalent plane graph with only hexagonal and pentag-onal faces. Let B = (V (B), E(B)) be an elementary circuit in Γ. The circuit B partitionsthe sphere into two open hemispheres. A patch Π of Γ is constructed by deleting all ver-tices, edges and faces interior to one of the hemispheres and replacing them by a single facebounded by B. Our notational convention Π = (VΠ, EΠ, FΠ, B) should be interpreted asfollows: (VΠ, EΠ) is the graph remaining after the deletion; FΠ is the collection of faces

E-mail addresses: jegraver@syr.edu (Jack E. Graver), cgraves@uttyler.edu (Christina M. Graves)

Copyright c© 2010 DMFA Slovenije

110 Ars Math. Contemp. 3 (2010) 109–120

of Γ remaining after the deletion. Hence, Π̂ = (VΠ, EΠ, F̂Π), where F̂Π is the collec-tion of faces FΠ along with the outside face, is a plane graph. The circuit B is called theboundary of the patch. Graphite patches are defined to be plane graphs with all hexagonalfaces save one outside face and with all vertices on the boundary of the outside face havingdegree 2 or 3 and with all other (internal) vertices having degree 3. While investigatinggraphite patches, Guo, Hansen and Zheng [6] produced two nonisomorphic patches withthe same boundary code, that is the same sequence of vertex degrees around the boundary.A fullerene or graphite patch that is not uniquely determined by its boundary is said to beambiguous.

When the boundary curve of the Guo, Hansen and Zheng ambiguous patches is tracedin the hexagonal tessellation of the plane or on a fullerene it intersects itself, that is, itsboundary is not an elementary circuit. In fact, it was shown in [5] that an ambiguousgraphite patch must triple cover some hexagon when projected onto the hexagonal tessel-lation of the plane. Our patches have elementary circuits for boundaries when drawn onthe fullerene. However, the boundary curve of a fullerene patch with only hexagonal facesmay have self intersections when projected onto the hexagonal tessellation of the plane. Inspite of this, it seems to be the case that fullerene patches with only hexagonal faces or atmost one pentagonal face cannot be ambiguous. But this has not been proved. In this paper,we show that ambiguous fullerene patches containing at most one pentagonal face, if theyexist, must be relatively large - roughly as large or larger than the smallest patch of thefullerene containing two pentagonal faces. In the papers [1], by Brinkmann, Caporossi andHansen, [2], by Brinkmann, Friedrichs and von Nathusius, and [3], by Brinkmann, Graverand Justus, it is shown that the number of faces in an ambiguous patch containing at mostone pentagonal face is uniquely determined.

The smallest example of an ambiguous fullerene patch is due to Endo and Kroto [4]and is pictured in Figure 1. These two patches have the same bounding circuit and containtwo pentagons. However, the relative positions of the pentagons are different in these twopatches. In fact, most of the patches enclosing two or more pentagonal faces are ambiguous.

Figure 1: The Endo-Kroto ambiguous patches.

Given any patch consisting of two non-adjacent pentagons joined by a simple path ofhexagons, we may construct a different patch with the same boundary. This generalizationof the Endo-Kroto construction is illustrated in Figure 2.

Let Π = (VΠ, EΠ, FΠ, B) be a patch of a fullerene Γ. If f is a face of Π, we denote theboundary circuit of f by B(f). The vertices of the boundary, B, of Π have degree 2 or 3 inΠ and one easily sees that every patch of size greater than 1 admits at least two boundaryvertices of degree 3. The paths in the circuit B joining two consecutive vertices of degree3 are called the segments of B. The proof of the following lemma is straight forward.

J. E. Graver and C. Graves: Fullerene Patches I 111

Π Π∗

Figure 2: Two ambiguous patches using the generalized Endo-Kroto construction.

Lemma 1.1. Let Π = (VΠ, EΠ, FΠ, B) be a patch of size at least 2. The path S =x0, e0, x1, . . . , ek, xk+1 is a segment of B if and only if S is a connected component ofB ∩B(f) for some face f of Π.

A face f is called an i-face if B∩B(f) consists of a single segment and that segment haslength i. By the distance between two faces in a fullerene we mean the distance betweenthe corresponding vertices in the dual graph. By the spread of a fullerene we mean theminimum distance between pentagonal faces taken over all pairs of pentagonal faces. Bya disk of radius r, we mean a patch consisting of all faces of distance r or less from acentral face (the center of the disk) as long as its boundary is an elementary circuit; seeFigure 3. For any patch, its diameter is the maximum of the distances (measured in theentire fullerene) between pairs of faces in the patch. Finally, the width of a patch is thediameter of the disk of least diameter containing it.

Figure 3: Three disks of radius 4.

The main result of this paper is:

Theorem 1.2. Ambiguous disks contain at least two pentagonal faces. In particular, alldisks with diameter less than the spread of the fullerene are unambiguous.

As a simple corollary of the main result we have:

112 Ars Math. Contemp. 3 (2010) 109–120

Corollary 1.3. In a fullerene, any patch with width less than the spread of the fullerene isunambiguous.

2 Proof of the main resultWe start this section by proving several basic lemmas. All patches that we investigate willbe considered as stand-alone patches. Hence they can be extended or altered even thoughthe extension or alteration cannot be carried out in the fullerene from which they came. Forthe patch Π = (VΠ, EΠ, FΠ, B), p(Π) will denote the number of pentagonal faces in thepatch and si, for i = 1 to 5, will denote the number of its i-faces.

Lemma 2.1. The number of pentagonal faces in a patch Π, p(Π), is given by p(Π) =6 + s1 − s3 − 2s4 − 3s5.

Proof. Let Π = (VΠ, EΠ, FΠ, B) be a patch in the fullerene Γ = (V,E, F ). Denote thenumber of vertices on the boundary of Π that have degree j (j = 2, 3) in Π by dj . Summingthe vertex degrees, we have 3|VΠ| − d2 = 2|EΠ| or 6|VΠ| − 4|EΠ| = 2d2. Summing theface degrees, we have 6h+5p(Π)+|B| = 2|EΠ|, where h denotes the number of hexagonalfaces of Π. Combining these two equations gives:

6|VΠ| − 6|EΠ| = 2d2 − (6h + 5p(Π) + |B|) = d2 − d3 − 6h− 5p(Π).

The number of face of the plane graph Π̂ is h + p(Π) + 1 and we then have 6|F̂Π| =6h + 6p(Π) + 6. Adding this and the previous equation gives:

6|VΠ| − 6|EΠ|+ 6|F̂Π| = d2 − d3 + p(Π) + 6.

By Euler’s formula, the left-hand side equals 12; so, we have p(Π) = 6− d2 + d3. Finally,one easily writes d2 and d3 in terms of s1, . . . , s5: d3 =

∑5i=1 si while d2 =

∑5i=1(i−1)si

giving

p(Π) = 6 +5∑

i=1

si −5∑

i=1

(i− 1)si = 6−5∑

i=1

(i− 2)si

The next lemma then follows at once.

Lemma 2.2. If Π and Π∗ are a pair of ambiguous patches then p(Π) = p(Π∗).

By a side of a patch Π we mean a set of boundary faces between two i-faces with i ≥ 3.Note that s3 + s4 + s5 is then the number of sides of Π. We define the length of a sidein terms of the distance between the end faces in the dual. Specifically, if all of the facesof a side have bounding segments of length 2 and if there are ` − 1 of them, we say theside has length ` or is an `-side. We will adopt the convention that a bounding 4-face isa side of length 0; this will eliminate the need for special cases of the formulas that wewill develop later. (5-faces will soon be eliminated from consideration, hence we adoptno special convention for them.) See Figure 4. A side is an h|k-side (h and k positiveintegers) if when rotating in the clockwise direction its first h− 1 faces and last k− 1 facesare 2-faces and they are separated by a single 1-face.

J. E. Graver and C. Graves: Fullerene Patches I 113

Side of length 3 1-side0-side

4|2-side

Figure 4: Side lengths.

A disk of radius r that consists solely of hexagons will be called an H-disk. A diskof radius r that contains exactly one pentagonal face is a P-disk. Notice that the only H-disk of radius 0 consists of a single hexagon, and the only P -disk of radius 0 is a singlepentagon.

Proposition 2.3. An H-disk with radius r greater than or equal to 1 has six r-sides. AP -disk with radius r greater than or equal to 1 has either

(ia) five r-sides,(ib) four r-sides and one t-side with r < t ≤ 2r or(ii) five r-sides and one h|k-side with h + k ≤ r.

Furthermore, the boundary faces of an H- or P -disk of radius r are precisely the faces ofdistance r from the center face of the disk.

Proof. The only H- and P -disks of radius 1 are easily constructed and are pictured inFigure 5. It is clear that these disks all satisfy the conclusions of the proposition. (Thesimplest P -patch of type (ii) is pictured in Figure 7.)

Figure 5: One H-disk and two P -disks of radius one.

Let Π be an H- or P -disk of radius r. We first show that all of the boundary faces area distance r from the center. Suppose that f is a boundary face of Π at a distance r′ from

114 Ars Math. Contemp. 3 (2010) 109–120

the center where r′ < r. Since, f is a boundary face, there must be a face f ′ of Γ not in Πbut adjacent to f . But then the distance of f ′ from the center is less than or equal to r and,by the definition of a disk, f ′ must be a face of the patch; contradiction.

We proceed by induction on the radius: assume that Π is an H- or P -disk of radiusr > 1 and that all H-disks and P -disks of radius less than r are accurately described bythe proposition. Delete all boundary faces of Π and any other faces that are a distance rfrom the center face c to get Π′. Since the faces of Π′ are all the faces of Π at a distance ofr − 1 or less from c and the faces of Π include all faces of Γ at a distance of r or less fromc, the faces of Π′ are all faces of Γ at a distance of r − 1 or less from c. Thus Π′ is an H-or P -disk of radius r − 1 and, by induction, it has the boundary structure described above.

Now consider a face f of Γ that is a distance r from c. It must be a face of Π and itmust share an edge with a boundary face of Π′. Hence, adding all of the faces of Γ not inΠ′ but adjacent to a boundary face of Π′, reconstructs Π. In reconstructing Π, we observethat when adding only hexagons to the boundary of an `-side produces an ` + 1-side andadding only hexagons to the boundary of an h|k-side produces another h|k-side.

Thus, if Π′ is an H-disk of radius r − 1, adding only hexagons creates another H-diskΠ with six r-sides. If Π′ is a P -disk, we may add only hexagons and we see immediatelythat: if Π′ is of type (ia) with five r−1-sides, Π is a P -disk of type (ia) with five r-sides; ifΠ′ is of type (ib) with four r− 1-sides and one t′-side (r− 1 < t′ < 2r− 2), Π is a P -diskof type (ib) with four r-sides and one t = t′ + 1-side (r < t < 2r − 1 < 2r); finally if Π′

is of type (ii) with five r− 1-sides and one h|k-side (h + k ≤ r− 1), Π is a P -disk of type(ii) with five r-sides and one h|k-side (h + k ≤ r − 1 ≤ r).

The only remaining case is that of adding a boundary of hexagons and exactly onepentagon to an H-disk of radius r − 1. If the pentagon is adjacent to the central edge of asegment of length 3, the result is a P -disk of type (ib) with four r-sides and one 2r-side.If the pentagon is not adjacent to the central edge of a length 3 segment, then it will havejust one edge on the boundary of the enlarged disk. Number the faces on the side of theenlarged disk that includes the pentagon f1, . . . , fr+1, where f1 and fr+1 have boundingsegments of length 3. If the pentagon is face fh+1, the P -disk resulting from adding thebounding faces will be of type (ii) with five r-sides and one h|k-side where k + h = r.

Corollary 2.4. H-disks are unambiguous.

Proof. When an H-disk is projected onto the hexagonal tessellation of the plane its bound-ary curve is an elementary circuit. Hence it is unambiguous.

To prove that P -disks are unambiguous, we are going to need to know just whichboundary curves are possible for certain classes of hexagonal patches - patches with onlyhexagonal faces. By Lemma 2.1, when a hexagonal patch has s1 = s4 = s5 = 0 it hasexactly six sides. If the lengths of these sides are a, b, c, d, e, f in cyclic order (counter-clockwise), we say the patch has side parameters, or simply parameters, (a, b, c, d, e, f).If a hexagonal patch has s1 = s5 = 0, s4 ≥ 1, then s3 = 6 − 2s4 and it will have 6 − s4

sides of positive length. By our convention, each 4-face is a side of length 0. Hence sucha patch can also be described by six side parameters (a, b, c, d, e, f) where non-adjacentvariables could take on the value 0; for example, the patch with parameters (a, 0, c, d, e, f)where a, c, d, e, f > 0 is a 5-sided hexagonal patch with the a and c sides separated by a4-face.

J. E. Graver and C. Graves: Fullerene Patches I 115

Lemma 2.5. Let Π be a hexagonal patch with s1 = s5 = 0 and side parameters (a, b, c, d,e, f). Then there are no consecutive sides of length 0 and the following three equalitiesmust hold: a + b = d + e, b + c = e + f and c + d = f + a.

Proof. If we project Π onto the hexagonal tessellation, then the boundary of the patchcan be represented in the dual (triangular) tessellation by the circuit joining consecutivebounding faces. In this representation, a side of length x is represented as straight lines ofx dual edges; these straight lines make 120◦ angles at 3-faces and 60◦ angles at 4-faces.Since s1 = s5 = 0 and s3 + 2s4 = 6 (by Lemma 2.1), we see that Π has at most three4-faces. If Π has three 4-faces, then Π is a “triangular” patch. It is clear that this trianglemust be an equilateral triangle and that all sides of Π are the same length. Thus, the sideparameters of Π are (a, 0, a, 0, a, 0) which do satisfy the above equalities. (See Figure 6.)

d

d

b=0

b+c

a+b

f

dc

bbb

a

e+f

d+e

b+c

a+b

f

ee e dc

bbb

a

e=0

e=0

a

c

d=a

f=ca=c+d

e=c

c

d d

c

f=0 0

b=0b=0

d=0f=0

c=a

e=a

a

Figure 6: Hexagonal Patch Boundaries.

If Π has exactly two 4-faces, called f1 and f2, then it must have two 3-faces, g1 andg2. If f1 and f2 are separated by only 2-faces, then projecting the patch onto the hexagonaltessellation has as its diagram a trapezoid with sides of length (a, 0, c, d, e, 0) where a isthe side between f1 and f2. It is apparent that sides a and d are parallel and that b + c =c = e = e+f . We can enlarge the patch by extending sides c and e to sides of length c+d.This new patch is an equilateral triangle, and thus a+b = a = e+d and a+f = a = c+d.On the other hand, if f1 and f2 are separated by 2-faces and one 3-face, then projecting Πonto the hexagonal tessellation yields a parallelogram with side parameters (a, 0, c, d, 0, f)which must satisfy a = d and c = f giving a + f = d + c, a + b = a = d = e + d andb + c = c = f = e + f .

If Π has exactly one 4-face, then projecting Π onto the hexagonal tessellation yields apatch with side parameters (a, b, c, d, 0, f) where sides f and c are parallel and sides a andd are parallel. By extending sides a and c by b hexagons, we get a parallelogram whichmust satisfy a+b = d and b+c = f . It follows that a+b = d = e+d and b+c = f = e+f .Adding a + b = d and f = b + c gives a + b + f = b + c + d and canceling the b givesc + d = f + a.

Finally, if Π has no 4-faces, then Π is a patch with side parameters (a, b, c, d, e, f)

116 Ars Math. Contemp. 3 (2010) 109–120

where sides a and d are parallel, sides b and e are parallel, and sides f and c are parallel.By extending sides a and c by d hexagons, and extending sides f and d by e hexagons, wecreate a parallelogram (see Figure 6). This parallelogram gives a+b = d+e, b+c = e+fand a + b + e + f = b + c + d + e directly and a + f = c + d by canceling the b and e inthe last equation.

If a P -disk Π were ambiguous, there exists a patch Π∗ with the same boundary. Sowe introduce a class of patches that will include all of these possible alternatives. By anr-patch, we mean a patch of one of the following two types. An r-patch of the first typeis a patch with s1 = s4 = s5 = 0, s3 = 5 and with 4 sides of length r and one sideof length t (t ≥ r). That is, an r-patch of the first type has side parameters (r, r, r, r, t),where t ≥ r. An r-patch of the second type is a patch with s4 = s5 = 0, s1 = 1 ands3 = 6, with five r-sides and one h|k-side. That is, an r-patch of the second type hasside parameters (r, r, r, r, r, h|k), where h and k are positive integers. It is an immediatecorollary of Lemma 2.1 that an r-patch contains exactly one pentagon. One easily checksthat there are only two 1-patches: the two P -disks of radius 1 with parameters (1, 1, 1, 1, 1)and (1, 1, 1, 1, 2). These are both r-patches of the first type and are pictured in Figure 5.The simplest r-patch of the second type has parameters (2, 2, 2, 2, 2, 1|1) and is pictured inFigure 7; note that it is a P -disk of radius 2.

Figure 7: The r-patch of the second type with parameters (2, 2, 2, 2, 2, 1|1).

The following lemma gives another property of r-patches — namely that we can deleteall bounding faces of an r-patch (r > 1) and still get a legitimate patch. It will always bepossible to remove all of the faces on a side unless one of those faces actually meets theboundary in two distinct segments — its removal would then leave a “disconnected” patch.It will always be possible to remove all of the faces on the boundary unless two faces ondifferent sides share an edge. See Figure 8.

Lemma 2.6. Let Π = (VΠ, EΠ, FΠ, B) be an r-patch r > 1.

i. Let f be a face of Π. Then, B(f) ∩B consists of at most one path.

ii. Let f and g be faces of Π that are not consecutive as one cyclically traverses theboundary. Then, B(f) ∩B(g) = ∅.

J. E. Graver and C. Graves: Fullerene Patches I 117

Figure 8: Impossible r-patch boundaries.

Proof. (i) Assume there exists a face f in Π consisting of 2 disjoint segments, a cut-face.Clearly both segments of f have length 2 or less. If both segments have length 2, then bothfaces adjacent to f are also cut-faces. Extending to the left, the string of cut-faces mustterminate. Terminating with a single face is impossible since this is an r-patch and such aface would be a 4-face or 5-face. Hence the leftmost cut-face must meet two faces on itsleft and must have at least one bounding segment of length 1. If f meets two faces on itsright, then it would have two bounding segments of length 1. If f meets only one face onits right then the right-most cut face in the string of cut-faces to the right must also have atleast one bounding segment of length 1. Hence this patch must have at least two boundingsegments of length 1 contradicting the assumption that it is an r-patch.

(ii) Consider any r-patch Π and suppose that it has a pentagonal 3-face f on its bound-ary. If the patch is of type (r, r, r, r, t), replacing f by a hexagon will result in a hexagonalpatch with boundary lengths (r, 0, r, r, r, t), (r, r, 0, r, r, t) or (r, r, r, r, 0, t); all of whichviolate the conditions of Lemma 2.5. If the patch is of type (r, r, r, r, r, h|k), replacing fby a hexagon will result in a hexagonal patch with boundary lengths (r, 0, r, r, r, r, h|k),(r, r, 0, r, r, r, h|k) or (r, r, r, r, r, 0, h|k). Now we can eliminate the segment of length 1in the h|k-side by attaching an h× k parallelogram of hexagons (see the left-hand diagramin Figure 9, also see Figure 10). The resulting six-sided hexagonal patches will then haveboundary lengths (h + r, 0, r, r, r, r + k), (h + r, r, r, 0, r, r + k) or (h + r, r, r, r, r, k); allof which violate the conditions of Lemma 2.5. Hence Π has no pentagonal 3-face f on itsboundary.

Let f and g be adjacent, non-consecutative boundary faces - a cut-pair. Since they arenon-consecutive, the face labeled y in the right-hand diagram of Figure 9 must belong tothe patch. If neither of the faces labeled x and z belong to the patch, then y would be ahexagonal 4-face, a pentagonal 3-face or a cut face. None of these is possible in an r-patch.Hence we may assume that z belongs to the patch. We may also assume without loss ofgenerality that f and g are the leftmost cut-pair and so x is also in the patch. By a similarargument v and one of u and w on the right must belong to the patch - say w. Hence gis a 1-face. If u is a face of the patch, then f is also a 1-face. If u is not a face of thepatch, then v and w form a cut pair and the right-most cut pair in this chain of cut-pairsmust also include a 1-face. Hence this patch must have at least two 1-faces contradictingthe assumption that it is an r-patch.

The core of a patch Π is the plane subgraph obtained by removing all the boundaryfaces of Π. By the construction of P -disks in Lemma 2.3, it is clear that a P -disk of radiusr has either an H-disk or a P -disk of radius r− 1 as its core. By Lemma 2.6, all r-patcheshave a legitimate patch as their core. We use this fact in carrying out the next step to

118 Ars Math. Contemp. 3 (2010) 109–120

r

r

rr

r

hh

k

kf

g

x

y

z

u

v

w

Figure 9: Diagrams for the proof of Lemma 5

eventually show that the only r-patches are the P -disks. The next step is to show that theparameters of any r-patches actually match the parameters of some P -disk.

Lemma 2.7. Let Π be an r-patch of the first type with parameters (r, r, r, r, t) Then t ≤ 2rand

i. if t = 2r, Π has a pentagon as the central face of the t-side;

ii. if t < 2r, Π has only hexagons on the boundary.

Proof. Let Π have parameters (r, r, r, r, t). In the proof of the previous lemma, we showedthat a pentagon in the boundary of an r-patch cannot be a 3-face. Suppose that the ith face(0 < i < r) in the first r-side is a pentagon. Replace this face by a hexagon. This newhexagonal patch has parameters (i, r− i, r, r, r, t); which is impossible since i + (r− i) 6=r + r. We get the same kind of contradiction if we assume that any face on any of ther-sides are pentagons. Now assume that the ith face (0 < i < t) in the t-side is a pentagon.Again, we replace the pentagon by a hexagon to get a hexagonal patch with parameters(r, r, r, r, i, t− i). We must have r +r = r + i and r +r = i+(t− i). This is possible onlyif i = r and t = 2r. Thus, the only way a pentagon can be on the boundary is if t = 2r andthe central face on the t-side is a pentagon.

To prove that t < 2r when Π has only hexagonal boundary faces, we’ll go by inductionon r. This inequality holds for r = 1 since the only 1-patch with only hexagonal boundaryfaces has parameters (1, 1, 1, 1, 1). Let Π be an r-patch with parameters (r, r, r, r, t) wherer > 1 and with only hexagons on its boundary. We can then remove the boundary andconsider its core. If a `-side has only hexagons, removing them produces an (` − 1)-side.Thus, the core must be an (r−1)-patch with side parameters (r−1, r−1, r−1, r−1, t−1).By induction, we have that (t− 1) ≤ 2(r − 1) or t < 2r.

Lemma 2.8. Let Π be an r-patch of the second type with parameters (r, r, r, r, r, h|k).Then h + k ≤ r and

i. if h + k = r, then Π has a pentagon as the face with bounding segment of length 1;

ii. if h + k < r, then Π has only hexagons on the boundary.

Proof. Let Π have parameters (r, r, r, r, r, h|k). Suppose that the ith face in the h|k-side,i < h, is a pentagon. We replace the pentagon by a hexagon to get a hexagonal patch with

J. E. Graver and C. Graves: Fullerene Patches I 119

parameters (r, r, r, r, r, i, (h−i)|k). If we attach the hexagons of a (h−i)×k parallelogram(see Figure 10), we get a 6-sided hexagonal patch with parameters (r, r, r, r, i+k, r+h−i).By Lemma 2.5, we have (i + k) + (r + h− i) = 2r; from which we conclude that we musthave h + k = r. Considering another adjacent pair, we have (r + h − i) + r = 2r; fromwhich we conclude that i = h. Thus the only case in which one of the faces on the h|k-sidecould be a pentagon is the case where h + k = r and the pentagonal face is the face withboundary segment of length 1.

r ih-i k r

r i

h-i

k

r

r i+k

r+h-i

Figure 10: Diagrams for the proof of Lemma 2.8

Next suppose that the ith face in first r-side is a pentagon. Replace it in the patch by ahexagon. This results in a hexagonal patch with parameters (i, r − i, r, r, r, r, h|k). Fillingin to eliminate the length 1 segment then yields a 6-sided hexagonal patch with parameters(i + h, r − i, r, r, r, r + k); which is impossible since (r − i) + r = r + (r + k) > 2r.We get the same kind of contradiction if we assume that any face on any of the r-sides arepentagons. Thus, if Π has a pentagon on the boundary, h + k = r and the pentagon is theface with bounding segment of length 1.

To prove that h + k < r when Π has only hexagonal boundary faces, we proceed byinduction on r. One easily checks that there are no 1-patches of this second type and theonly 2-patch of the second type is pictured in Figure 7 and has parameters (2, 2, 2, 2, 2, 1|1).Hence, the lemma holds for all r-patches with r ≤ 2. Let Π be an r-patch with parameters(r, r, r, r, r, h|k) for r > 2 and with only hexagons on its boundary. We can remove theboundary and consider its core. If an r-side has only hexagons, removing them produces an(r − 1)-side; if an h|k-side has only hexagons, removing them produces another h|k-side.Thus, the core must be an (r − 1)-patch with sides (r − 1, r − 1, r − 1, r − 1, r − 1, h|k).By induction, we see that h + k ≤ r − 1 or h + k < r.

This final proposition concludes our proof of the main result.

Proposition 2.9. P -disks are unambiguous.

Proof. This proof will proceed by induction on r. It is easy to check that the only twoP -disks of radius 1 are unambiguous. Hence we assume that Π is a P -disk of radius r > 1and that P -disks of radius r − 1 are unambiguous. Let Π∗ be another patch with the sameboundary as Π. By Proposition 2.3, Π∗ must be an r-patch with the same parameters as Π.We will show that Π∗ is in fact the same patch as Π.

For the first case, we assume that Π has parameters (r, r, r, r, t) with t < 2r. ByLemma 2.7, we see that Π and Π∗ have only hexagons on their boundaries. Stripping off

120 Ars Math. Contemp. 3 (2010) 109–120

the hexagonal boundary faces of both Π and Π∗ yields in each case a core that is an (r−1)-patch with parameters (r−1, r−1, r−1, r−1, t−1). By Proposition 2.3, there is a P -diskwith these parameters and by the induction hypothesis it must be the only (r − 1)-patchwith these parameters. Hence, Π and Π∗ have the same P -disk as core. There is only oneway to add hexagons around the boundary of a P -disk, so Π∗ must be the same patch as Π.

Now, assume that Π has parameters (r, r, r, r, 2r). By Lemma 2.7, both Π and Π∗ havea pentagon as the central face of the 2r-side. Thus the core of both Π and Π∗ is the H-diskof radius r − 1. Replacing the boundary faces, we see that Π∗ = Π.

If Π has parameters (r, r, r, r, r, h|k) with h+k < r, both Π and Π∗ have only hexagonson the boundary. Thus, the core of Π is the P -disk with radius r − 1 and parameters(r − 1, r − 1, r − 1, r − 1, r − 1, h|k). Since the core of Π∗ is an r-patch with the sameparameters, it is by induction the same P -disk as the core of Π. Again, replacing theboundary faces, we see that Π∗ = Π.

Finally, assume Π has parameters (r, r, r, r, r, h|k) with h + k = r. By Lemma 2.8,Π and Π∗ must have a pentagon on the boundary and that pentagon is the boundary facewith bounding segment of length 1. Thus, the core of both Π and Π∗ is the H-disk ofradius r − 1. As above Π and Π∗ have the same core and the same boundary faces and aretherefore equal.

References[1] G. Brinkmann, G. Caporossi and P. Hansen, A survey and new results on computer enumeration

of polyhex and fusene hydrocarbons, J. Chem. Inf. Comput. Sci. 43 (2003), 842–851.

[2] G. Brinkmann, O. Delgado Friedrichs and U. von Nathusius, Numbers of faces and boundaryencodings of patches, Graphs and discovery, 27–38, DIMACS Ser. Discrete Math. Theoret.Comput. Sci., 69, AMS, Providence, RI, 2005.

[3] G. Brinkmann, J. E. Graver and C. Justus, Numbers of Faces in Disordered Patches, J. Math.Chem. 45 (2009), 263–278.

[4] M. Endo and H. W. Kroto, Formation of carbon nanofibers, J. Phys. Chem. 96 (1992), 6941–6944.

[5] J. E. Graver, The (m,k)-patch boundary code problem, MATCH 48 (2003), 189–196.

[6] X. Guo, P. Hansen and M. Zheng, Boundary Uniqueness of Fusenes, Tech. Report G-99-37,GERAD, 1999.